What is the Correct Approach to Solving the Deuteron Transcendental Equation?

AI Thread Summary
The discussion centers around solving the deuteron transcendental equation from Krane's nuclear physics textbook, specifically the equation k1 cot(k1R) = -k2. Participants calculate the parameter b, with values around 0.4856 and 0.4627, depending on the reduced mass and energy used. There is a focus on converting MeV to Joules and ensuring the correct application of physical constants. Discrepancies in calculated values for b highlight the importance of using accurate mass values for protons and neutrons. Ultimately, the goal is to confirm that V0 approximates 36 MeV based on the solutions derived.
James_1978
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Homework Statement
I believe I solved the transcendental equation but the plot does not make sense.
Relevant Equations
##k_{1} \cot{k_{1}R} = -k_{2}##
##k_{1} = \frac{\sqrt{2m(E+V_{o})}}{\hbar}##
##k_{2} = \frac{\sqrt{-2mE}}{\hbar}##
##x = -\tan{bx}##
##x = \sqrt{\frac{-(V_{o} + E)}{E}}##
Dear Forum,

I am trying to solve a problem (4.6) from the introductory nuclear physics textbook by Krane. The problem is as follows:
Solving the deuteron using the radial equations gives the transcendental function,

##k_{1} \cot{k_{1}R} = -k_{2}##

Were

##k_{1} = \frac{\sqrt{2m(E+V_{o})}}{\hbar}##

And

##k_{2} = \frac{\sqrt{-2mE}}{\hbar}##

That gives the relations between and R. Show that this equation can be written in the form,

##x = -\tan{bx}##

Where

##x = \sqrt{\frac{-(V_{o} + E)}{E}}##

Evaluate the parameter b for R = 2fm. Note that is the reduced mass. Solve the transcendental equation.

When rearranging we get ##b## as.

##b = \frac{\sqrt{-2mE}}{\hbar}*R##

For the reduced mass ##m = \frac{1.67x10^{-27}}{2} kg##
For ##\hbar = 1.054x10^{-34} J-s##
For ##E = -2.22 MeV##

We are suppose to see that when solving the transcendental equation we get ##V_{o} = 36 MeV##. However we must have something wrong because the function does not clearly show how you infer the ##V_{o} = 36 MeV##. Any help is appreciated.
 
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How much value of b you get ? Please let me know it for checking your result.
 
Last edited:
Dear Anuttarasmmyak

Here is what I got for b

##b = \frac{\sqrt{-2*\frac{1.67x10^{-27}}{2}*-2.22*1.602x10^{-13}}}{1.054x10^{-34}}##

Where ##b =2.3155x10^{14} m^{-1}##

Or with R I get ##b*R = 0.4856## I get ##V_{o} = 36##. I think this is correct. Just wanted to make sure.

1676640075854.png
 
b has no physical dimension.
 
Yes. I saw that. b*R is unit-less. My mistake.
 
Is MeV translated to MKSA Joule ?
 
Yes. I think you are asking in that I multiplied MeV*1.602x10^-13 to convert MeV to Joules. Is that what you are asking?
 
And you say R is 2 fm.
 
Yes, I use 2x10-15 m.
 
  • #10
James_1978 said:
Or with R I get
So you say b=0.4856.
 
  • #11
Yes. That is what I got.
 
  • #13
anuttarasammyak said:
So you say b=0.4856.
James_1978 said:
Yes. That is what I got.
I got a different value for ##b##. I calculated ##b=0.4627##.
 
  • #14
What did you use for E and mass of proton?
 
  • #15
I first calculated ##b## using the numbers you used and got a different answer. So I looked up the mass of a proton and neutron and found the reduced mass (##1.673\times 10^{-27}~\rm kg##) and used that to get the number above. Either way, I didn't get the value for ##b## you found. In both cases I used ##E_1 = -2.22~\rm MeV##.

In fact, I don't get the same values for the calculations you showed in post #3 for ##b## (really ##b/R##) or ##b R## (really ##b##). Moreover, your two values don't make sense if you're using ##R=2~\rm fm##.
 
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